Two bodies of same shape, same size and same radiating power have emissivities 0.2 and 0.4 The ratio of their temperature is .

Options

A. `sqrt(3) : 1`
B. `sqrt(2) : 1`
C. `1 : sqrt(5)`
D. `1 : sqrt(8)`

Answer

B

Explanation

Radiating power of a body of area `A_(1)`, emissivity `e_(1)` and surface temperature `T_(1)` is
`P_(1)=sigmae_(1)T_(1)^(4)A_(2)" "...(i)`
Similarly, radiating power of a body of area `A_(2)`, emissitivity `e_(2)` and surface temperature `T_(2)` is
`P_(2)=sigmae_(2)T_(2)^(4)A_(2)" "...(ii)`
Given `P_(1)=P_(2)" and "A_(1)=A_(2)`
`sigmae_(1)T_(1)^(4)A_(1)=sigmae_(2)T_(2)^(4)A_(2)`
`implies" "((T_(1))/(T_(2)))^(4)=((e_(2))/(e_(1)))((A_(2))/(A_(1)))`
`(T_(1))/(T_(2))=((e_(2))/(e_(1)))^(1//4).1" "(becauseA_(1)=A_(2))`
Here, `e_(1)=0.4,e_(2)=0.16`
`:.(T_(1))/(T_(2))=((0.4)/(0.16))^(1//4)=((4)/(1))^(1//4)=((2^(2))/(1^(1)))^(1//4)`
`(T_(1))/(T_(2))=((2)/(1))^(1//2)=sqrt((2)/(1))`
`T_(1):T_(2)=sqrt(2):1`